Continuity of $f(x)= \frac{1}{3-x}$ at a $x=4$ point using $\varepsilon$-$\delta$

Question

Let $f(x)= \frac{1}{3-x}$ for all $x \in \mathbb{R}$. I want to show that $f$ is continuous at $x=4$.

We have $| f(x) - f(c) | = \left|\frac{1}{3-x}-\frac{1}{-1}\right| $. Suppose also that $|x-4| < \delta $. I'm not sure what to do next.

Answer 1

If $|x-4|<\frac12$, then $4-\frac12<x<4+\frac12$, and therefore $1-\frac12<x-3<1+\frac12$; in particular, $|x-3|>\frac12$. So, for a given $\varepsilon>0$, take $\delta=\min\{\frac\varepsilon2,\frac12\}$. Then$$|x-4|<\delta\implies\frac{|x-4|}{|x-3|}<\frac1{1/2}\frac\varepsilon2=\varepsilon.$$

Answer 2

$$| f(x) - f(c) | = \left|\frac{1}{3-x}-\frac{1}{-1}\right|=\left|\frac{4-x}{3-x}\right|$$

Notice that

$$\left|\frac{4-x}{3-x}\right|<\left|4-x\right|\quad\text{for } x<2 \quad\text{and}\quad4<x$$

This inequality is easy to prove.

So $$\left|\frac{4-x}{3-x}\right|<\left|4-x\right|<\delta$$

Therefore, just take $\delta=\min\{\varepsilon,2\}$